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Straight Lines - 1 - JEE MCQ


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30 Questions MCQ Test Mathematics (Maths) for JEE Main & Advanced - Straight Lines - 1

Straight Lines - 1 for JEE 2024 is part of Mathematics (Maths) for JEE Main & Advanced preparation. The Straight Lines - 1 questions and answers have been prepared according to the JEE exam syllabus.The Straight Lines - 1 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Straight Lines - 1 below.
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Straight Lines - 1 - Question 1

The point which divides the joint of (1, 2) and (3,4) externally in the ratio 1 : 1.

Detailed Solution for Straight Lines - 1 - Question 1

x1 = (1*3 - 1*1)/(1-1)
y1 = (1*4 - 1*2)/(1-1)
Hence the denominator becomes zero, here we can say that any point cannot be divided in the ratio of 1:1 externally.

Straight Lines - 1 - Question 2

The distance of the point (α,β) from X axis is

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Straight Lines - 1 - Question 3

The line through the points (a , b) and (- a, - b) passes through the point

Detailed Solution for Straight Lines - 1 - Question 3

Slope of line passing through (a,b) and (−a,−b) is given by (b+b)/(a+a) = b/a
So equation of line passing is given by (using slope point form)
y−b = b/a(x−a)
⇒ ay − ab = bx − ab
⇒ ay = bx
Clearly the point (a2,ab) lie on the above line

Straight Lines - 1 - Question 4

The equation y−y1 = m (x−x1), m ∈ R, represents all lines through the point (x1,y1) except the line

Straight Lines - 1 - Question 5

Slope of any line parallel yo X axis is

Straight Lines - 1 - Question 6

The straight lines x + y = 0 , 3x + y – 4 = 0 , x + 3y – 4 = 0 form a triangle which is

Detailed Solution for Straight Lines - 1 - Question 6

Explanation:- x+y=0 .....  (i) 
3x+y−4=0 ..... (ii) 
x+3y−4=0 ....... (iii)
Solving lines (i) and (ii), we get
−x=−3x+4
⟹ x=2, y=−2
∴(i) and (ii) intersect at A=(2,−2)
Solving lines (ii) and (iii), we get
−3x+4 = (4−x)/3
⟹ −9x+12 = 4−x
⟹x=1, y=1
∴(ii) and (iii) intersect at B=(1,1)
Solving lines (i) and (iii), we get
-x = (4−x)/3
⟹ −3x = 4−x
⟹ x=−2, y=2
So, AB,BC,AC form a triangle ABC
Now, AB = [(1−2)2 + (1+2)2]1/2 = (10)1/2
BC = [(−2−1)2 + (2−1)2]1/2 = (10)1/2
AC = [(−2−2)2 + (2+2)2]1/2 = (32)1/2
​∴ AB=BC
Since, two sides of a triangle are equal then the triangle formed by A,B,C is isosceles triangle.

Straight Lines - 1 - Question 7

The vertex A of a triangle ABC is the point (-2, 3) whereas the line perpendicular to the sides AB and AC are x – y – 4 = 0 and 2x – y – 5 = 0 respectively. The right bisectors of sides meet at P(3/2 , 5/2) . Then the equation of the median of side BC is

Detailed Solution for Straight Lines - 1 - Question 7

Given vertex of ΔABC A(−2,3)
Equation of side AB perpendiuclar to x−y−4=0 is 
AB:x+y−λ=0
Side AB passes through point A(−2,3) 
−2+3−λ=0
λ=1
Hence AB:x+y−1=0−−−−(1)
Equation of side AC perpendiuclar to 2x−y−5=0 is 
AC:x+2y−λ=0
Side AC passes through point A(−2,3) 
−2+6−λ=0
λ=4
Hence AC:x+2y−4=0−−−−(2)
The right bisectors of all sides are meeting at point P(3/2 , 5/2)
Right bisector from point B on side AC is perpendicular
Hence Equation of right bisector perpendicular to x+2y−4=0 is 
2x−y−λ=0
Here right bisector is passing through point P (2 × 3/2 − 5/2​)λ = 0
=> 6/2 − 5/2 = λ
λ=1/2 
Equation of right bisector BD be 
2x−y−1/2 =0
4x−2y−1=0−−−−(3)
Here On solving equation of AB and BD we get point of intersection i.e. point B
4(1−y)−2y−1=0
4−4y−2y−1=0
−6y=−3
y = 1/2
​x = 1 − y = 1− 1/2 = 1/2
Point B(1/2,1/2)
Right bisector from point A on side BC is passing through point P
Hence equation of AD from point A and P
y−3 = {[5/2−3](x+2)/(3/2+2)}
y−3 = −1/7(x+2)
Slope of AD is mAD = −1/7
Hence slope of side BC perpendicular to AD is −1/7mBC = −1
mBC = 7
Equation of side BC from point B(1/2,1/2) with slope mBC=7
y − 1/2 = 7(x−1/2)
2y−1=14x−7
14x−2y−6=0

Straight Lines - 1 - Question 8


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Straight Lines - 1 - Question 9


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Straight Lines - 1 - Question 10


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Straight Lines - 1 - Question 11


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Straight Lines - 1 - Question 12


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Straight Lines - 1 - Question 13


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Straight Lines - 1 - Question 14

The ends of the base of an isosceles triangle are at (2, 0) and (0, 1) and the equation of one side is x = 2 then the orthocentre of the triangle is

Detailed Solution for Straight Lines - 1 - Question 14


Straight Lines - 1 - Question 15


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Straight Lines - 1 - Question 16


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Straight Lines - 1 - Question 17

B & C are fixed points having co-ordinates (3, 0) and (- 3, 0) respectively. If the vertical angle BAC is 90º, then the locus of the centroid of the  D ABC has the equation :

Detailed Solution for Straight Lines - 1 - Question 17

Straight Lines - 1 - Question 18


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Straight Lines - 1 - Question 19


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Straight Lines - 1 - Question 20

Given A(1, 1)  and AB is any line through it cutting the x-axis in B. If AC is perpendicular to AB and meets the y-axis in C, then the equation of locus of mid- point P of BC is

Detailed Solution for Straight Lines - 1 - Question 20



Straight Lines - 1 - Question 21


Detailed Solution for Straight Lines - 1 - Question 21


Straight Lines - 1 - Question 22

If 2a2 – 7ab – ac + 3b2 – 2bc – c2 = 0 then the family of lines ax + by + c = 0 are either concurrent at the point P(x1, y1) or at the point Q(x2, y2). Find the distance of the origin from the line passing through the points P and Q, the distance being measured parallel to the line 3x – 4y = 2.

Detailed Solution for Straight Lines - 1 - Question 22

Straight Lines - 1 - Question 23

Three straight lines are drawn through a point P lying in the interior of the  D ABC and parallel to its sides. The areas of the three resulting triangles with P as the vertex are  s1, s2 and  s3. The area of the triangle in terms of  s1, s2 and  s3 is :    

Detailed Solution for Straight Lines - 1 - Question 23



Straight Lines - 1 - Question 24


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Straight Lines - 1 - Question 25

An equilateral triangle ABC is inscribed in the parabola y = x2 and one of the side of the equilateral triangle has the gradient 2. If the sum of x-coordinates of the vertices of the triangle is a rational in the form p/q  where p and q are coprime, then find the value of (p + q).

Detailed Solution for Straight Lines - 1 - Question 25


Straight Lines - 1 - Question 26

Line x + 2y = 4 is translated by √5 units closer to the origin and then rotated by angle tan–1 1/2 in the clockwise direction about the point where the shifted line cuts the x-axis. Find the distance of new line from M(3, 3).

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Straight Lines - 1 - Question 27

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Straight Lines - 1 - Question 28


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Straight Lines - 1 - Question 29


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Straight Lines - 1 - Question 30


Detailed Solution for Straight Lines - 1 - Question 30


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